Magnitude of the tension on the ends of the clothesline?

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The discussion focuses on calculating the tension in a horizontal clothesline tied between two poles, with a 2-kilogram mass causing a 3-meter sag. Participants clarify that the tension is not simply the weight of the mass but involves both vertical and horizontal components due to the angle of the sag. The vertical components from both ends of the clothesline must equal the weight of the mass, leading to a more complex calculation than initially assumed. The correct approach involves using trigonometric functions to resolve the components of tension accurately. Ultimately, understanding the vector nature of tension is crucial for solving the problem correctly.
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A horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 2 kilograms is tied to the middle of the clothesline, it sags a distance of 3 meters.

What is the magnitude of the tension on the ends of the clothesline?

Isnt the tension just 2g Newtons? or for just one end of the clothesline g Newtons? The website where i enter these problems is not validating these answers. I don't know why. I also did the x componets, the y componets, and the negative of all of that, the answers are never valid. I have no clue why.
 
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You must also account for the horizontal componet of force. Draw some triangles.
 
the horizontal componet and the vertical componet adds up to the weight, which is 2(9.81)N. isn't that the tension?
 
No, the vertical component of each of the two ends of the clothesline is equal to .5*1*9.8 N. Consider that the direction of the tension is not directly up, but up and sideways because it must be in the same direction as the line. The net tension at one end will therefore be the vector sum of the horizontal and vertical tensions.
 
shouldnt the vertical componet of each pole be: 1*9.8*sin[tan^(-1)(3/7)]?

how did you get .5*1*9.8?
 
The vertical components of each of the two must add up to the weight of the hanging mass, which is downward.
 
ok, i got it, thanks
 
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